Question

The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

Hint:

Use: \[ \alpha = \frac{R_2 - R_1}{R_1 \Delta T} \] Works directly when reference temperature is known.

Answer 1

Calculation of Temperature Coefficient of Resistivity
We are given:
- Resistance at \( T_1 = 20^\circ \text{C} \): \( R_1 = 1.05 \, \Omega \) - Resistance at \( T_2 = 100^\circ \text{C} \): \( R_2 = 1.38 \, \Omega \) The resistance of a metal varies with temperature according to:
\[ R_T = R_0 \left[ 1 + \alpha (T - T_0) \right] \] where \( \alpha \) is the temperature coefficient of resistivity and \( R_0 \) is the resistance at reference temperature \( T_0 \). Here, \( R_0 = R_1 \) at \( T_0 = 20^\circ \text{C} \).
Step 1: Write the equation for \( R_2 \)
\[ R_2 = R_1 \left[ 1 + \alpha (T_2 - T_1) \right] \] Substitute the values:
\[ 1.38 = 1.05 \left[ 1 + \alpha (100 - 20) \right] \] \[ 1.38 = 1.05 \left[ 1 + 80 \alpha \right] \]
Step 2: Solve for \( \alpha \)
\[ \frac{1.38}{1.05} = 1 + 80 \alpha \] \[ 1.3143 \approx 1 + 80 \alpha \] \[ 80 \alpha = 0.3143 \] \[ \alpha = \frac{0.3143}{80} \approx 0.00393 \, \text{°C}^{-1} \]
Step 3: Summary
- The temperature coefficient of resistivity of the metal is:
\[ \boxed{\alpha \approx 3.93 \times 10^{-3} \, \text{°C}^{-1}} \]
This positive value indicates that the metal's resistance increases with temperature.